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Transform the equation

Question:

Transform the equation $2 x^{2}+y^{2}-4 x+4 y=0$ to parallel axes when the origin is shifted to the point (1, -2).

Solution:

Let the new origin be (h, k) = (1, -2)

Then, the transformation formula become:

$x=X+1$ and $y=Y+(-2)=Y-2$

Substituting the value of x and y in the given equation, we get

$2 x^{2}+y^{2}-4 x+4 y=0$

Thus

$2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0$

$\Rightarrow 2\left(X^{2}+1+2 X\right)+\left(Y^{2}+4-4 Y\right)-4 X-4+4 Y-8=0$

$\Rightarrow 2 X^{2}+2+4 X+Y^{2}+4-4 Y-4 X+4 Y-12=0$

$\Rightarrow 2 X^{2}+Y^{2}-6=0$

$\Rightarrow 2 X^{2}+Y^{2}=6$

Hence, the transformed equation is $2 X^{2}+Y^{2}=6$